# Quadratic Relations Review

### By: Sarah Krigos

## Introduction

## Multiplying Binomials

Multiplying binomials is the simplest thing you can do in this unit. All you do is expand and simplify. Use the distributive property to expand binomials and then collect like terms to simplify. Here's an example:

(x+2) (x+5)

-there are a few simple steps in solving this

step 1:

(x+2) (x+5)

multiply the first number or variable in the first bracket by both in the second bracket:

(x *x) (x*5)

x² + 5x

step 2:

multiply the second variable or number in the first bracket by both in the second bracket:

(2*x) (2*5)

x² + 5x +2x +10

step 3:

collect like terms

x² + __5x +2x__ +10

step 4:

write out the answer

x² + 7x + 10

now were going to solve this one all together

(x-4) (x+7)

= x² + 7x - 4x - 28

=x² + 3x - 28

## Common Factoring

Factoring is the opposite of expanding. If every term of a polynomial is divisible by the same constant, the constant is called a common factor. common factoring is also very simple. all it requires is your knowledge of your times tables. This requires only two simple steps in order to solve. Here's an example:

8x-12y

step 1:

find the greatest common factor

-although we know that this can be factored by more than 1 number we always have to go with the greatest. If we were to factor this by 2 because 2 can go into 8 and 12 it wouldn't be completely factored because 2 is not the greatest common factor and it would be considered wrong. that's why we always go with the greatest common factor which in this case is 4.

step 2:

divide the greatest common factor into both terms and then write it out

8/4 = 2 12/4=3

4 (2x-3y)

now were going to do one all together

10a-14b+6c

GCF: 2

2(5a-7b+3c)

## Factoring by Grouping

In the last part we looked at common factoring. however in reality we are never going to always get questions were everything can be completely factored by the same number so that's why were going to be looking at factoring by grouping which is basically factoring the equation in two parts instead of one. There are a few simple steps in solving this. Here's an example:

x² +7x+6

step 1:

find the product of the last term and the sum of the second.

-in this case we know that product is multiplication and sum is addition. so we are basically trying to find 2 numbers that multiply to 6 and add to 7

which in this case is 1 and 6

step 2:

write the numbers out in the bracket

(x+1) (x+6)

*always remember to include the variable when you right out the numbers*

now were going to try a few all together:

1. x² -8x+12

P:12

S: -8

(x-6) (x-2)

2. x² -5x-14

P: -14

S: -5

(x-7) (x+2)

3. x² +3x+5

P:5

S:3

not factorable

## Factoring Complex Trinomials

In the last slide we looked a little at factoring by grouping through product and sum. now were going to go more into that and the break down of how to factor complex trinomials. Here's an example:

2x² + 3x - 5

step1:

Find 2 numbers that make the product of the first number multiplied by the last number and that make the sum of the second number.

-whenever there is a number in front of the x we always multiply the first number by the last number. when it is just x it doesn't matter if you multiply it because we consider x as just 1 so when you multiply it, it wont change. therefore now instead of our product being -5 and the sum being 3 the product is now -10 and the sum is still 3.

in this case the numbers are 5 and -2

step 2:

expand it with the numbers

2x² + 5x - 2x -5

step 3:

factor the first two terms and the last two terms

**2x² + 5x** __- 2x -5__

x(2x +5) -1(2x+5)

step 4:

simplify

-put the numbers that are in the brackets first because it is there twice

x(**2x +5**) -1(**2x+5**)

(2x+5)

-put the left over numbers in brackets as well

**x**(2x +5)** -1**(2x+5)

(2x+5) (x-1)

now were going do some more all together

5x²+12x+4

P:20

S:12

5x²+10x +2x+4

5x(x+2) +2(x+2)

(x+2) (5x+2)

## solving quadratic equations

So far we've looked at how to expand and factor however we haven't looked at how to solve these equations. Here's an example of how to solve one:

x² +3x+2=0

step 1:

find the product and sum

P:2

S:3

=2,1

step 2:

write it out

(x+1) (x+2)=0

step 3:

make the numbers in the brackets into equations

x+1=0

x+2=0

step 4:

solve

x=-1

x=-2

now lets do one all together

x² -7x-18=0

P: -18

S: -7

(x-9) (x+2)=0

x-9=0

x+2=0

x=9

x=-2

## Completing the square

Completing the square is what allows you to change an equation into vertex form. completing the square is a vertex simple process that requires few steps to do. Here's an example:

x² +6x -2

step1:

block off the first two terms

(x² +6x) -2

step 2:

Factor out the "A"

x(x+6x)-2

step 3:

add zero: divide the middle term by 2 and then square it)

(x² +6x +9-9)-2

step 4:

bring out the negative

(x²+6x+9)-9-2

step 5:

simplify

(x+3)²-11

now lets do one all together

3x² +12x +15

(3x²+12x)+15

3(x²+4x) +15

3(x²+4x+4-4) +15

3(x²+4x+4)-__12__+15 *note that -4 turned into negative 12 because we multiplied it by 3 before we took it out of the bracket*

3(x+2)²+3

## The Quadratic Formula

The quadratic formula is a formula used to solve for x. It is not so hard to do just difficult to often memorize lets take a look at it:

ax² + bx + c = 0

The quadratic formulaIt looks complicated but its actually quite simple. Since you'll be given the values of a b and c all you have do is solve for x. Here's an example:

a=1

b=-1

c=-6

x² + x - 6= 0

step1:

sub in the variables

x= - (-1) +/- √(-1)² - 4 (1) (6) /2(1)

step2:

solve the part that is to be square rooted

x= - (-1) +/- **√(-1)² - 4 (1) (6) /2(1)**

x= 1 +/- √25 /2

step3:

add and then subtract the numbers and divide both answers by 2

x= 1+/- 5 / 2

x= 1+5/2

x= 1-5/2

x=-2 x=3

now lets do one all together:

4x² +16x +12

x= -(16) +/- √(16)² - 4 (4) (12) /2(4)

x=-(16) +/- √(256) - (192) / 8

x= -(16) +/- √64 / 8

x= -(16) +/- 8 / 8

x= -(16) + 8 / 8

x= -(16) - 8 / 8

x= -1

x= -3